Coalpail.com Forum

Lightning wrote:Larry, pardon me for saying so but I think you being a bit stubborn at this point lol. I have to agree,, the fan doesn't produce any heat. Hell the stove doesn't produce any heat either. It's merely a heat transfer station. The only thing producing heat is the coal.

The elevated volume of air moving across the stove's surface is causing better heat transfer efficiency. I believe this was the original poster's intention.

Lightning wrote:Is this a trick question?
Volts x amps equals watts
Watts define BTU output.
If volts drop, more amperage would be needed to maintain heat output. Fan or no fan.

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tony17112acst wrote:I'm the OP. *bows*

My assumption in my question is that the same amount of energy is used inside the stove with and without the fan.

Please consider the original question with say, an electric resistance coil heater set up inside of the stove. This heater can only put out a certain number of BTU's ...say 50,000 BTU's ...and no more!

So I ask you recent posters: Will more heat transfer to the room by pointing a fan on the stove that contains this 50,000 BTU electric heater? You cannot claim that more heat will come out of the electric heater inside the stove (because it's impossible). The clear answer is "yes." It is "yes" because a fan will cause a lower temp on the outside of the stove and more heat will transfer to the room and not out the pipe/chimney because of the principle that exists that heat transfers through a medium more quickly as the temperature difference is greater on either side of it.

EFFICIENCY IS A GOOD THING. BUT...
Solid fuel appliances only work because of the The Chimney Effect - by losing a certain amount of heat into the chimney, they make the gases there rise and so safely evacuate smoke and pull fresh air onto the fuel to make it burn. An appliance with a very high efficiency is losing less heat into the chimney and, unless its chimney is perfect, it may encounter problems in generating enough draught to make it burn effectively, or safely evacuate smoke, it may be problematic in lighting-up from cold and may not keep the waste gases hot enough to prevent tar from condensing inside the flue. This is why appliances with the very highest efficiencies, some above 90% (net), almost invariably require an electric chimney fan.

MEASURING EFFICIENCY and OUTPUT
It is not normally practical to directly measure the heat output of a solid fuel appliance, instead, samples of the waste flue gas are taken - usually using a flue gas analyser the chemical composition of the gas indicates how efficiently the fuel has been burned and its temperature indicates how much of the heat generated has been captured and how much is lost into the chimney.

Given the temperature of and percentage of carbon dioxide (CO2) in the exiting flue gas, the, moderately accurate, Siegert's formula gives the efficiency from:

100 - ( (MeanFlueTemp ºC - MeanRoomTemp ºC) x (A1 / CO2%) )

(Where A1 is: Anthracite=0.683, Coke=0.290, Bituminous Coal=0.672, Lignite=1, Peat=0.7, Dry Wood=0.650)

The heat output in kW is then calculated from:

(Efficiency x Potential heat in kW x Fuel burned in kg) / Burn time in hours

...using data on the Fuel Properties page

freetown fred wrote:As in, one mans relevant is another mans regurgitate

freetown fred wrote:As in, one mans relevant is another mans regurgitate

PJT wrote:...is it over?...

freetown fred wrote:WAIT----was that the fat lady I heard singing???